📢 Binomial Theorem – Class 11 Mathematics 📚

🚀 Master the Binomial Theorem for Board & JEE Exams! 🚀

The Binomial Theorem is a fundamental topic in Class 11 Mathematics, widely used in algebra, combinatorics, and calculus. It helps in expanding expressions of the form (a + b)ⁿ efficiently, making it essential for solving complex problems in CBSE Board Exams, JEE Mains, and Advanced.

Binomial theorem is used to find the expansion of two terms hence it is called the Binomial Theorem. Binomial theorem is a fundamental principle in algebra that describes the algebraic expansion of powers of a binomial. According to this theorem, the expression (a + b)ⁿ where a and b are any numbers and n is a non-negative integer. It can be expanded into the sum of terms involving powers of a and b.

It gives an expression to calculate the expansion of algebraic expression (a+b)ⁿ. The terms in the expansion of the following expression are exponent terms and the constant term associated with each term is called the coefficient of terms.

Binomial Theorem Statement

The Binomial Theorem provides a systematic way to expand the powers of a binomial expression. Binomial theorem for the expansion of (a+b)ⁿis stated as,

(a + b)ⁿ = ⁿC₀aⁿb⁰ + ⁿC₁a^n-1b¹ + ⁿC₂a^n-2b² + …. + ⁿC_ra^n-rb^r + …. + ⁿC_na⁰bⁿ

where n > 0 and the ⁿC_k is the binomial coefficient.

Binomial Expansion Formula

Binomial theorem formula gives the expansion of the algebraic identities in the form of a series. This formula is used to find the expansion up to n terms of the (a+b)ⁿ. The binomial expansion of (a + b)ⁿ can easily be represented with the summation formula.

Binomial Theorem Formula for the expansion of (a + b)ⁿ is,

(a + b)ⁿ= ∑ⁿ_rⁿC_r a^n-rb^r

where,

n is a positive integer,
a, b are real numbers, and 0 < r ≤ n

We can easily find the expansion of the various identities such as (x+y)9, (x+12)¹², and others using the Binomial Theorem Formula. We can also find the expansion of (ax + by)ⁿ using the Binomial Theorem Formula,

Expansion formula for (ax + by)ⁿ is,

(ax + by)ⁿ = ∑ⁿ_rⁿC_r (ax)^n-r(by)^r

where 0< r ≤ n.

Also using combination formula we know that,

Binomial Expansion Formula

ⁿC_r = n! / [r! (n – r)!]

Binomial Theorem Proof

We can easily proof Binomial Theorem Expansion using the concept of Principle of Mathematical Induction

Let’s take x, a, n ∈ N, and then the Binomial Theorem says that,

(x+y)ⁿ = ⁿC₀ xⁿy⁰ + ⁿC₁ x^n-1y¹ + ⁿC₂ x^n-2 y² + … + ⁿC_n-1 x¹y^n-1 + ⁿC_n x⁰yⁿ

Proof:

For n = 1, we get

(x +y)¹ = x +y which is true.

For n = 2,

(x +y)² = (x + y) (x +y)
⇒ (x +y)² = x² + xy + xy + y² (using distributive property of multiplication over addition)
⇒ (x +y)² = x² + 2xy + y²

which is also true.

Thus, theorem is true for n = 1 and n = 2.

Let’s take a positive integer k.

(x+y)^k = ^kC₀ x^ky⁰ + ^kC₁ x^k-1y¹ + ^kC₂ x^k-2 y² + … + ^kC_k-1 x¹y^k-1 + ^kC_k x⁰y^k

Now consider the expansion for n = k + 1

(x + y) ^k+1 = (x + y) (x + y)^k

⇒ (x + y) ^k+1 = (x + y) (x^k + ^kC₁ x^k-1y¹ +^kC₂ x^k-2 y² + … + ^kC_r x^k-ry^r +….+ y^k)

⇒ (x + y) ^k+1 = x^k+1 + (1 + ^kC₁)x^ky + (^kC₁ + ^kC₂) x^k-1y² + … + (^kC_r-1 + ^kC_r) x^k-r+1y^r+ … + (^kC_k-1 + 1) xy^k + yk⁺¹

⇒ (x + y) ^k+1 = x^k+1 + ^k+1C₁x^ky + ^k+1C₂ x^k-1y² + … + ^k+1C_r x^k-r+1y^r + … + ^k+1C_k xy^k + y^k+1

[As we know, ⁿC_r + ⁿC_r-1 = ⁿ⁺¹C_r]

This result is true for n = k+1.

Thus, by concept of mathematical induction, the result is true for all positive integers ‘n’. Proved.

Binomial Expansion

Binomial Theorem is used to expand the algebraic identity (x + y)ⁿ. Hence it is also called the binomial expansion. The binomial expansion of (x + y)ⁿ with the help of the binomial theorem is,

(x+y)ⁿ = ⁿC₀ xⁿy⁰ + ⁿC₁ x^n-1y¹ + ⁿC₂ x^n-2 y² + … + ⁿC_n-1 x¹y^n-1 + ⁿC_n x⁰yⁿ

Using this expansion we get each term in the expansion of (x+y)ⁿ

Some special cases from the binomial theorem can be written as:

(x + y)ⁿ = ⁿC₀ xⁿ + ⁿC₁ x^n-1 by+ ⁿC₂ x^n-2 y² + … + ⁿC_n-1 x y^n-1 + ⁿC_n xⁿ
(x – y)ⁿ = ⁿC₀ xⁿ – ⁿC₁ x^n-1 by + ⁿC₂ x^n-2 y² + … + (-1)ⁿ ⁿC_n xⁿ
(1 – x)ⁿ = ⁿC₀ – ⁿC₁ x + ⁿC₂ x² – …. (-1)ⁿ ⁿC_n xⁿ

Also, ⁿC₀ = ⁿC_n = 1

However, there will be (n + 1) terms in the expansion of (a + b)ⁿ.

Properties of Binomial Theorem

There are various properties related to the binomial theorem, some of those properties are as follows:

Number of Terms: In the binomial Expansion of (x + y)ⁿ using the binomial theorem, there are n+1 terms and coefficients.
First and last terms in the binomial expansion of (x + y)ⁿ are xⁿ and yⁿ, respectively.
General Term: In the binomial expansion of (x + y)ⁿ the general term can be represented as T(r + 1) and is given by T(r + 1) = ⁿC_r × x^(n-r) × y^r.
Pascal’s Triangle: The binomial coefficients in the expansion are arranged in Pascal’s triangle which is the pattern of number where each number is the sum of the two numbers above it.

Specific Values: When n is a non-negative integer, the expansion simplifies for specific values of n:
- (a + b)⁰ = 1
- (a + b)¹ = a + b
- (a + b)² = a² + 2ab + b²
- (a + b)³ = a³ + 3a²b + 3ab² + b³
- and so on…
Binomial Coefficients Symmetry: In the binomial expansion of (x + y)ⁿ, the r^th term from the end is the (n – r + 2)^th term from the beginning.
If n is even, then in the expansion of (x + y)ⁿ, the middle term is ((n/2) + 1). If n is odd, then the middle terms are ((n + 1)/2) and ((n + 3)/2) in the expansion of (x + y)ⁿ.

Binomial Expansion for Negative Exponent

Binomial theorem is also used for finding the expansion of the identities which have negative exponents. The coefficients terms in the negative expansion are similar in magnitude to the terms in the corresponding positive exponent.

Some of the simplified expansions of the negative exponents which are widely used are,

(1 + x)^-1 = 1 – x + x² – x³ + x⁴ – x⁵ + …….
(1 – x)^-1 = 1 + x + x² + x³ + x⁴ + x⁵ + …….
(1 + x)^-2 = 1 – 2x + 3x² – 4x³ + ……..
(1 – x)^-2 = 1 + 2x + 3x² + 4x³ + ……..

These expansions are easily proved using binomial expansion and replacing (+) with (-)

Applications of Binomial Theorem

The Binomial Theorem has various applications it is used for a variety of the purposes such as,

Finding Remainder in the division of very large numbers.
Finding Last Digits of a Number
Checking the Divisibility

Finding Remainder using Binomial Theorem

This can easily be understood with the help of the following example.

Example.1: Find the remainder when 2⁹⁷ is divided by 15

Solution:

(2⁹⁷ / 15) = [2(2⁴)²⁴ / 15)]

= [2(15 + 1)²⁴ / 15]

As each term of the expansion of (1 + 15)²⁴ contains 15 except the first term which is only 1.

Thus, Remainder when 2⁹⁷ is divided by 15 is 1.

Example.2: Find the remainder when 7¹⁰³ is divided by 25.

Solution:

(7¹⁰³/ 25) = [7(49)⁵¹ / 25)] = [7(50 − 1)⁵¹/ 25]

= [7(25K − 1) / 25] = [(175K – 25 + 25−7) / 25]

= [(25(7K − 1) + 18) / 25]

∴ The remainder = 18

Example.3: If the fractional part of the number (2⁴⁰³ / 15) is (K/15), then find K.

Solution:

(2⁴⁰³ / 15) = [2³(2⁴)¹⁰⁰ / 15]

= 8/15 (15 + 1)¹⁰⁰ = 8/ 15 (15λ + 1) = 8λ + 8/15

∵ 8λ is an integer, fractional part = 8/15

So, K = 8.

Finding the Last Digits of a Number

How to find the last digit of an expansion can be understood using the example,

Example.1: Find the last digit of (7)¹⁰

Solution:

(7)¹⁰ = (49)⁵ = (50-1)⁵

(50-1)⁵ = ⁵C₀ (50)⁵ − ⁵C₁ (50)⁴ + 5C₂ (50)3 − ⁵C₃ (50)² + ⁵C₄ (50) − ⁵C₅

= ⁵C₀ (50)⁵ − ⁵C₁ (50)⁴ + 5C₂ (50)3 − ⁵C₃ (50)² + 5(50) − 1

= ⁵C₀ (50)⁵ − ⁵C₁ (50)⁴ + 5C₂ (50)3 − ⁵C₃ (50)² + 249

A multiple of 50 + 249 = 50K + 249

Thus, the last digit is 9.

Example.2 : Find the last two digits of the number (13)¹⁰

Solution :

(13)¹⁰= (169)⁵= (170 − 1)⁵

= 5C₀ (170)⁵− 5C₁(170)⁴+ 5C₂(170)³− 5C₃(170)²+ 5C₄(170) − 5C₅

= 5C₀(170)⁵− 5C₁(170)⁴+ 5C₂(170)³− 5C₃(170)²+ 5(170) − 1

A multiple of 100 + 5(170) – 1 = 100K + 849

∴ The last two digits are 49.

Example.3 : The digit in the units place of the number 183! + 3¹⁸³.

Solution:

⇒ 3¹⁸³= (3⁴)⁴⁵.3³

⇒ unit digit = 7 and 183! ends with 0

∴ The units digit of 183! + 3¹⁸³is 7.

Relation between Two Numbers

Example.1 : Find the larger of 99⁵⁰ + 100⁵⁰ and 101⁵⁰

Solution :

101⁵⁰= (100 + 1)⁵⁰= 100⁵⁰+ 50 . 100⁴⁹+ 25 . 49 . 100⁴⁸+ …

⇒ 99⁵⁰= (100 − 1)⁵⁰= 100⁵⁰– 50 . 100⁴⁹+ 25 . 49 . 100⁴⁸− ….

⇒ 101⁵⁰– 99⁵⁰= 2[50 . 100⁴⁹+ 25(49) (16) 100⁴⁷+ …]

= 100⁵⁰+ 50 . 49 . 16 . 100⁴⁷+ … >100⁵⁰

∴ 101⁵⁰– 99⁵⁰> 100⁵⁰

⇒ 101⁵⁰> 100⁵⁰+ 99⁵⁰

Divisibility Test

Example.1 : Show that 11⁹+ 9¹¹ is divisible by 10.

Solution :

11⁹+ 9¹¹= (10 + 1)⁹+ (10 − 1)¹¹

= (9C₀. 10⁹+ 9C₁. 10⁸+ … 9C₉) + (11C₀. 10¹¹− 11C₁. 10¹⁰+ … −11C₁₁)

= 9C₀. 10⁹+ 9C₁. 10⁸+ … + 9C₈. 10 + 1 + 10¹¹− 11C₁. 10¹⁰+ … + 11C₁₀. 10−1

= 10[9C₀. 10⁸+ 9C₁. 10⁷+ … + 9C₈+ 11C₀. 10¹⁰− 11C₁. 10⁹+ … + 11C₁₀]

= 10K, which is divisible by 10.

Binomial Theorem for Any Index

Binomial Theorem for any index, including non-integer and negative indices, generalizes the familiar binomial expansion that applies to positive integer exponents.

This generalized form involves the use of binomial coefficients that are defined for any real number index using the concept of factorial functions extended to the gamma function for non-integer values.

Multinomial Theorem

We know that the binomial theorem expansion of (x + a)ⁿ is,

(x+a)ⁿ= ⁿ∑_rⁿC_r x^{n – r} a^r

Where n∈N

We can generalize this result to get the expansion of,

(x₁ + x₂+ … +x_k)ⁿ = ∑_{(r1 + r2 + …. + rk = n)} [n! / r₁!r₂!…r_k!] x₁^r1 x₂^r2 …x_kr^k

The general term in the above expansion is

[n! / r₁!r₂!…r_k!] x₁^r1 x₂^r2 …x_kr^k

The number of terms in the above expansion is equal to the number of non-negative integral solutions of the equation r₁+ r₂ + … + r_k = n

Each solution of this equation gives a term in the above multinomial expansion. The total number of solutions can be given by ^{n + k – 1}C_{k −1}.

Binomial Distribution and Binomial Coefficients

Binomial Distribution

The binomial distribution models the number of successes in a fixed number of independent and identical Bernoulli trials (experiments with two possible outcomes: success or failure). It’s characterized by two parameters: nn (the number of trials) and pp (the probability of success in each trial).

Hypergeometric Distribution

The hypergeometric distribution models the probability of obtaining a specific number of successes in a sample of a fixed size drawn without replacement from a finite population containing a known number of successes and failures.

Important Formulae:

The number of terms in the expansion of (x₁+ x₂+ … x_r)ⁿ is (n + r − 1)C_{r – 1}
The sum of the coefficients of (ax + by)ⁿ is (a + b)ⁿ

If f(x) = (a₀+ a₁x + a₂x²+ …. + a_mx^m)ⁿ then

(a) Sum of coefficients = f(1)
(b) Sum of coefficients of even powers of x is: [f(1) + f(−1)] / 2
(c) Sum of coefficients of odd powers of x is [f(1) − f(−1)] / 2

Binomial Theorem for Rational Index

The number of rational terms in the expression of (a^1/l + b^1/k )ⁿ is [n / LCM of {l,k}] when none of and is a factor of and when at least one of and is a factor of is [n / LCM of {l,k}] + 1 where [.] is the greatest integer function.

Example.1 : Find the number of irrational terms in (⁸√5 + ⁶√2)¹⁰⁰.

Solution :

T_{r + 1} = 100C_r (⁸√5)^{100 – r}. (⁶√2)^r= 100C_r. 5[(100 – r)/8] .2^r/6.

∴ r = 12,36,60,84

The number of rational terms = 4

The number of irrational terms = 101 – 4 = 97

Number of Terms in the Expansion

Number of Terms in the Expansion of (x₁ + x₂ + … + x_r)ⁿ

From the general term of the above expansion, we can conclude that the number of terms is equal to the number of ways different powers can be distributed to x₁, x₂, x₃…., x_n, such that the sum of powers is always “n”.

The number of non-negative integral solutions of

x₁+ x₂+ … + x_r= n is ^{n +r – 1}C_{r – 1}.

For example, the number of terms in the expansion of

(x + y + z)³ is ^{3 + 3 -1}C_{3 – 1}= ⁵C₂= 10

As in the expansion, we have terms such as

As x⁰y⁰ z⁰, x⁰ y¹ z², x⁰ y² z¹, x⁰ y³ z⁰, x¹ y⁰ z², x¹ y¹ z¹, x¹ y² z⁰, x² y⁰ z¹, x² y¹ z⁰, x³ y⁰ z⁰.

The number of terms in (x + y + z)ⁿ is ^{n + 3 – 1}C_{3 – 1}= ^{n + 2}C₂.

The number of terms in (x + y + z + w)ⁿ is ^{n + 4 – 1}C_{4 – 1} = ^{n + 3}C₃ and so on.

Important Sum of Binomial Coefficients

Some properties of combination which are widely used in simplifying binomial expansion are,

C₁ + C₂ + C₃ + C₄ + …….C_n = 2ⁿ
C₀ + C₂ + C₄ + …. = C₁ + C₃ + C₅ + ……. = 2^n-1
C₀ – C₁+ C₂ – C₃+ C₄ – C₅+ …. = 0
C₁ + 2C₂ + 3C₃ + 4C₄ + …….nC_n = n2^n-1
C₁ – 2C₂ + 3C₃ – 4C₄ + …….(-1)ⁿnC_n = 0
C₁² + C₂² + C₃² + C₄² + …….C_n² = (2n)! / (n!)²

Pascal’s Triangle Binomial Expansion

The number associated with the terms of the binomial expansion is called the coefficient of the binomial expansion. These variables can easily be found using Pascal’s Triangle or by using combination formulas.

Binomial Theorem Coefficients

After examining the coefficient of the various terms in the expansion of algebraic identities using the binomial theorem, we have observed a trend in the coefficient of the terms of the expansion. The trend is that the coefficient of the terms in the expansion of the binomial terms is directly similar to the rows of the Pascal triangle.

For example, the coefficient of each term in the expansion of the (x+y)⁴ is directly equivalent to the terms in the 4^th row of the Pascal Triangle.

Pascal Triangle

Pascal Triangle is an alternative method of the calculation of coefficients that come in binomial expansions, using a diagram rather than algebraic methods.

In the diagram shown below, it is noticed that each number in the triangle is the sum of the two directly above it. This pattern continues indefinitely to obtain coefficients of any index of the binomial expression.

When we observe the pattern of the coefficients of the expansion (a + b)ⁿ, Pascal’s triangle for the pattern of the coefficients of the expansion (a + b)⁷is shown in the figure below:

Thus, from the above diagram, the expansion of small powers of n (e.g. n=0, 1, 2, 3, 4, 5, 6, 7) can be calculated as:

(a + b)⁰ = 1

(a + b)¹= a + b

(a + b)² = a²+ 2ab +b²

(a + b)³ = a³ + 3a²b + 3ab² + b³

(a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴

(a + b)⁵ = a⁵ + 5a⁴b + 10a³b² + 10a²b³ + 5ab⁴ + b⁵

(a + b)⁶ = a⁶ + 6a⁵b + 15a⁴b² + 20a³b³ + 15a²b⁴ + 6ab⁵ + b⁶

(a + b)⁷ = a⁷ + 7a⁶b + 21a⁵b² + 35a⁴b³ + 35a³b⁴ + 21a²b⁵ + 7ab⁶ + b⁷

In this way, the expansion from (a + b)⁰ to (a + b)⁷ is obtained by the application of Pascal’s Triangle. But finding (a + b)¹⁵ is really a long process using Pascal’s triangle. So, here we use combinations formula.

Combinations

The combination formulas are also widely used to find coefficients of the various terms in the expansion of the binomial theorem.

The combination formula used for choosing r objects out of n total objects is widely used in the binomial expansion and it is defined as,

ⁿC_r = n! / [r! (n – r)!]

Also, some common combination formulas used in the Binomial Expansion are,

ⁿC_n = ⁿC₀ = 1
ⁿC₁ = ⁿC_n-1 = n
ⁿC_r = ⁿC_r-1

Solved Examples of Binomial Theorem

Example.1: Find the value of (x+y)² and (x+y)³ using Binomial expansion.

Solution:

(x+y)² = ²C₀ x²y⁰ + ²C₁ x^2-1y¹ + ²C₂ x^2-2 y²

⇒ (x+y)²= x² + 2xy + y²

And (x+y)³ = ³C₀ x³y⁰ + ³C₁ x^3-1y¹ + ³C₂ x^3-2 y² + ³C₃ x^3-3 y³

⇒ (x+y)³= x³ + 3x²y + 3xy² + y³

⇒ (x+y)³= x³ + 3xy(x+y) + y³

Example.2 : Find the expansion of (x + 5)⁶ using the binomial theorem.

Solution:

We know that here,

(a + b)ⁿ = ⁿC₀aⁿb⁰ + ⁿC₁a^n-1b¹ + ⁿC₂a^n-2b² + …. + ⁿC_ra^n-rb^r + …. + ⁿC_na⁰bⁿ

Thus, (x + 5)⁶ = ⁶C₀x⁶5⁰ + ⁶C₁ x^6-15¹ + ⁶C₂ x^6-25² + ⁶C₃ x^6-35³ + ⁶C₄ x^6-45⁴ + ⁶C₅ x^6-55⁵ + ⁶C₆ x^6-65⁶

= ⁶C₀x⁶ + ⁶C₁ x⁵5 + ⁶C₂ x⁴5² + ⁶C₃ x³5³ + ⁶C₄ x²5⁴ + ⁶C₅ x¹5⁵ + ⁶C₆ x⁰5⁶

= x⁶ + 30x⁵ + 375x⁴ + 2500x³ + 9375x² + 18750x + 15625

Example 3: Expand the binomial expression (2x + 3y)².

Solution:

(2x + 3y)²

= (2x)² + 2(2x)(3y) + (3y)²

= 4x² + 12xy + 9y²

Example 4: Expand the following (1 – x + x²)⁴

Solution:

Put 1 – x = y

Then,

(1 – x + x²)⁴ = (y + x²)⁴

= ⁴C₀y⁴(x²)⁰+ ⁴C₁y³(x²)¹+ ⁴C₂y²(x²)²+⁴C₃y(x²)³+⁴C₄(x²)⁴

= y⁴+ 4y³x²+ 6y²x⁴+ 4yx⁶+ x⁸

= (1 – x)⁴ + 4(1 – x)³x² + 6(1 – x)²x⁴ + 4(1 – x)x⁶ + x⁸

= 1 – 4x + 10x² – 16x³ + 19x⁴ – 16x⁵ + 10x⁶– 4x⁷ + x⁸

Example 5: Find the binomial expansion of (x³ + 1)³.

Solution:

We know that,

(x + y)ⁿ = ⁿC₀ xⁿy⁰ + ⁿC₁ x^n-1y¹ + ⁿC₂ x^n-2 y² + … + ⁿC_n-1 x¹y^n-1 + ⁿC_n x⁰yⁿ

(x³ + 1)³ = ³C₀ (x³)³1⁰ + ³C₁ (x³)²1¹ + ³C₂ (x³)¹1²+ ³C₃ (x³)⁰1³

(x³ + 1)³ = (1)x⁹ + (3)x⁶ + (3)x³ + (1)

(x³ + 1)³ = x⁹ + 3x⁶ + 3x³ + 1

Example.6: Expand: [x² + (3/x)]⁴, x ≠ 0

Solution:

[x² + (3/x)]⁴

Using binomial theorem,

[x² + (3/x)]⁴ = ⁴C₀ (x²)⁴ + ⁴C₁ (x²)³ (3/x) + ⁴C₂ (x²)² (3/x)² + ⁴C₃ (x²) (3/x)³ + ⁴C₄ (3/x)⁴

= x⁸ + 4 x⁶ (3/x) + 6 x⁴ (9/x²) + 4 x² (27/x³) + (81/x⁴)

= x⁸ + 12x⁵ + 54x² + (108/x) + (81/x⁴)

Example 7: Compute (98)⁵

Solution:

Let us write the number 98 as the difference between the two numbers.

98 = 100 – 2

So, (98)⁵ = (100 – 2)⁵

Using binomial expansion,

(98)⁵ = ⁵C₀ (100)⁵ – ⁵C₁ (100)⁴ (2) + ⁵C₂ (100)³ (2)² – ⁵C₃ (100)² (2)³ + ⁵C₄ (100) (2)⁴ – ⁵C₅ (2)⁵

= 10000000000 – 5 × 100000000 × 2 + 10 × 1000000 × 4 – 10 ×10000 × 8 + 5 × 100 × 16 – 32

= 10040008000 – 1000800032

= 9039207968

Example.8 : Find the number of terms in (1 + 2x +x²)⁵⁰

Solution :

(1 + 2x + x²)⁵⁰= [(1 + x)²]⁵⁰= (1 + x)¹⁰⁰

The number of terms = (100 + 1) = 101

Binomial Theorem – FAQs

What is Binomial Theorem?

Binomial theorem is the basic theorem which provides an expansion to algebraic identity such as (x+y)ⁿ

(x+y)ⁿ = ⁿC₀ xⁿy⁰ + ⁿC₁ x^n-1y¹ + ⁿC₂ x^n-2 y² + … + ⁿC_n-1 x¹y^n-1 + ⁿC_n x⁰yⁿ

What is General Term in Binomial Theorem?

General term is the binomial theorem expansion of (x+y)ⁿ is,

T_r+1 = ⁿC_r x^n-ry^r

What Is Constant Term in the Binomial Theorem?

Term in the binomial theorem expansion which is free from the variables is called the constant term.

What is the Binomial Theorem formula?

Binomial theorem is used to find the expansion of the algebraic terms of the form(x + y)ⁿ. The formula of the binomial expansion is,

(x + y)ⁿ = Σ_rⁿⁿC_rx^{n – r}· y^r

What are the number of terms in the expansion of (x + a)ⁿ + (x-a)ⁿ?

The number of terms in the expansion of (x + a)ⁿ + (x-a)ⁿ is given with the help of the formula,

If n is even (n+2)/2

If n is odd (n+1)/2

🔍 Applications of the Binomial Theorem

📌 Algebra & Number Theory – Polynomial expansion, finding coefficients
📌 Probability & Statistics – Combinations & binomial distributions
📌 Physics & Engineering – Approximation in equations of motion
📌 Computer Science & Coding – Algorithm efficiency & complexity analysis

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📢 Binomial Theorem – Class 11 Mathematics 📚

Table of Contents

Binomial Theorem Statement

Binomial Expansion Formula

Binomial Expansion Formula

Binomial Theorem Proof

Binomial Expansion

Properties of Binomial Theorem

Binomial Expansion for Negative Exponent

Applications of Binomial Theorem

Finding Remainder using Binomial Theorem

Finding the Last Digits of a Number

Relation between Two Numbers

Divisibility Test

Binomial Theorem for Any Index

Multinomial Theorem

Binomial Distribution and Binomial Coefficients

Binomial Distribution

Hypergeometric Distribution

Important Formulae:

Binomial Theorem for Rational Index

Number of Terms in the Expansion

Important Sum of Binomial Coefficients

Pascal’s Triangle Binomial Expansion

Binomial Theorem Coefficients

Pascal Triangle

Combinations

Solved Examples of Binomial Theorem

Binomial Theorem – FAQs

What is Binomial Theorem?

What is General Term in Binomial Theorem?

What Is Constant Term in the Binomial Theorem?

What is the Binomial Theorem formula?

What are the number of terms in the expansion of (x + a)n + (x-a)n?

🔍 Applications of the Binomial Theorem

📚 Why Choose Neeraj Anand’s Book?

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